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1 2 Algebraic Extensions 1. Let K be a field and E an extension of K. One writes this assumption in short as Let E=K be a field extension; and the word field is often omitted when it can be inferred from the context. An element of E is called algebraic over K if there exists a polynomial f.x / 0 in KŒX such that f. /D 0: If is not algebraic over K, we say that is transcendental over K. Remarks. (a) If K D and E D, the elements of E algebraic over K are called simply algebraic numbers, and the elements of E transcendental over K are called transcendental numbers. Example: WD 3p 2 is an algebraic number, since is a root of the polynomial X ŒX. (b) The set of algebraic numbers is countable (since ŒX is countable and any nonzero polynomial in ŒX has finitely many roots in ). Therefore the set of transcendental numbers must be uncountable. To actually be able to exhibit a transcendental number is a different (and much harder) matter. Theorem 1. Let M be a subset of containing 0 and 1. Any point z 2 algebraic over K WD.M [ M /. M is The proof will be given later in this chapter. But first we quote a famous result: Theorem 2 (Lindemann 1882). The number is transcendental. Corollary. The quadrature of the circle with ruler and compass is impossible. Proof. If it were possible, we would have 2 ; by Theorem 1 then would be algebraic, which by Lindemann s Theorem is not the case. Lindemann s Theorem can be proved using relatively elementary algebraic and analytic arguments, but the proof is on the whole quite intricate. We will go into it later on (Chapter 17).

2 16 2 Algebraic Extensions 2. Now we start our study of field theory with the following statement: F1. Let E=K be a field extension. If 2 E is algebraic over K, then K. /WK < 1: Proof. Suppose there exists a nonzero polynomial (1) f.x / D X n C a n 1 X n 1 CCa 0 2 KŒX such that f. /D 0; we have assumed without loss of generality that f is normalized (has leading coefficient 1). There exists a unique homomorphism of K-algebras ' from the polynomial ring KŒX into E such that '.X / D (see page 21); its image R D im ' E consists precisely of those elements of E that can be written as polynomial expressions g. / in with coefficients in K. But in writing such an expression we immediately see from the relation (2) n D.a n 1 n 1 CCa 1 C a 0 / that only terms of degree less than n are needed, so in fact (3) R Dfc 0 C c 1 CCc n 1 n 1 j c i 2 Kg: Thus, as a vector space over K, the dimension of R is at most n. Since R, being a subring of E, has no zero-divisors, a simple argument (given a bit further down) shows that R is actually a field. It follows that K. / R (using the definition of K. /), and therefore that R D K. /. From (3) we then get (4) K. / Dfc 0 C c 1 CCc n 1 n 1 j c i 2 Kg: In particular, (5) K. /WK n. F2. Let R be an integral domain (that is, a commutative ring with no zero divisors and with 1 0), and let K be a subfield of R. If R is finite-dimensional as a K-vector space, R is a field. Proof. For a given a 0 in R, consider the map h W R! R given by multiplication by a, namely, h.x/ D ax for all x in R. Thenh is an endomorphism (linear map) of the K-vector space R. Since R has no zero-divisors, h is injective. Because R is assumed finite-dimensional over K, it is also surjective. In particular, there exists b 2 R such that ab D 1. Remark. It can be proved in an analogous way that an integral domain that has finite cardinality is a field.

3 The minimal polynomial Let E=K be a field extension, and let 2 E be algebraic over K. Consider on the K-vector space K. / the endomorphism h defined by multiplication by. The minimal polynomial of h is called the minimal polynomial of over K, andwe denote it by MiPo K. /: This is the lowest-degree normalized polynomial in KŒX that has as a zero. (That there can be only one such polynomial is clear: if f;g are both normalized and of degree n, the degree of f g is less than n.)thedegreeoff D MiPo.K/ is also called the degree of over K, and is denoted by Œ WK. Example. Consider E D, K D and D e 2i=3.Then is a root of X 3 1. But X 3 1 D.X 1/g.X /, with g.x / D X 2 C X C 1; since 1, wehaveg. / D 0. Let f D MiPo K. /; we claim that f D g. Otherwise necessarily deg f<deg g, so f could only be of the form f.x / D X, which is impossible since. F3. Let E=K be a field extension and let 2 E be algebraic over K, of degree n WD Œ WK. The elements (6) 1; ; 2;:::; n 1 of E form a basis of K. / over K. In particular, (7) K. /WK D Œ WK D deg MiPo K. /: Proof. Let f.x / D X n CCa 1 X C a 0 the minimal polynomial of over K. We know that K. /WK ni see (5) in the proof of F1. There remains to show that 1; ; 2;:::; n 1 are linearly independent over K. Suppose there is a relation (8) Xn 1 c i i D 0 with c i 2 K: id0 Set g.x / WD P n 1 id0 c ix i.ifsomec i in (8) were nonzero, g.x / would be a nonzero polynomial in KŒX of degree less than n and vanishing at. Contradiction! 4. Let E=K be a field extension and assume 2 E is algebraic over K. Is it the case that any ˇ 2 K. / is also algebraic over K? Definition. An extension E=K is called algebraic if every element of E is algebraic over K. An extension E=K is called finite if E WK < 1. Remarks. = is a finite extension, since W D 2. The extension = is not algebraic; see Remark (b) in Section 2.1. An extension E=K is called transcendental if it is not algebraic.

4 18 2 Algebraic Extensions F4. If an extension E=K is finite, it is also algebraic; for each ˇ 2 E the degree Œˇ WK is a divisor of E WK. Proof. Let E=K be finite of degree n. Given ˇ 2 E, the n C 1 elements 1;ˇ; ˇ2;:::;ˇn of the n-dimensional K-vector space E are linearly dependent. Therefore there exist a 0 ; a 1 ;:::;a n 2 K, not all zero, such that a 0 1 C a 1ˇ CCa nˇn D 0: Thus ˇ is algebraic over K. ByF3,Œˇ WK D K.ˇ/ W K, andk.ˇ/ WK is a divisor of E WK by the degree formula (Chapter 1, F7). We now can easily answer in the affirmative the question asked at the beginning of this section. F5. Let E=K be a field extension. If 2 E is algebraic over K, the extension K. /=K is algebraic. Proof. If is algebraic over K, we know from F1 that K. /=K is finite. But every finite field extension is algebraic, by F4. Together, F1 and F4 afford the following criterion: F6. Let E=K be a field extension. An element of E is algebraic over K if and only if K. /=K is finite. Now it is a cinch to prove Theorem 1, which we can reformulate as follows: Theorem 1. Let M be a subset of containing 0 and 1. Let K D.M [ M /. The field extension M=K is algebraic. Proof. Take z 2 M. From F9 of Chapter 1 we know that K.z/WK < 1. ThenF6 says that z is algebraic over K. Remark. The converse of F4 is not true: Not every algebraic extension is finite. This will soon become obvious. In fact a counterexample comes up naturally in our context: If E D f0; 1g is the field of all numbers constructible from f0; 1g with ruler and compass, the field extension E= is algebraic but not finite. (With what we know so far this is not very easy to prove, but it s worth thinking about; see 2.5 in the Appendix.) Among algebraic extensions, finite extensions can be characterized thus: F7. Let E=K be a field extension. The following conditions are equivalent: (i) There are elements 1;:::; m of E, finite in number and algebraic over K, such that E D K. 1;:::; m/. (ii) E=K is finite.

5 Properties of algebraic extensions 19 Proof. (ii) ) (i) is clear; all we need to do is choose a basis 1;:::; m for E=K. Then we actually have E D K 1 CCK m, and by F4 all the i are algebraic over K. To show (i) ) (ii) we use induction over m. For m D 0 there is nothing to prove. Assume that (i) holds for some m 1 and set K 0 D K. 1;:::; m 1 /: Then E D K 0. m/. Since m is algebraic over K, itisa fortiori algebraic over the larger field K 0. By F1 this implies E W K 0 < 1. But by the induction hypothesis, K 0 =K is finite. The degree formula (Chapter 1, F7) then implies that E=K is finite. 5. Let E=K be a field extension. A subfield L of E containing K is called an intermediate field of the extension E=K. F8. Let E=K be a field extension. The subset F Df 2 E j is algebraic over Kg is an intermediate field of E=K. It is called the algebraic closure of K in E. In particular, the set of all algebraic numbers is a subfield of. Proof. Take ; ˇ 2 F. Consider the subfield K. ; ˇ/ of E. By F7 the extension K. ; ˇ/=K is finite (prove this again for practice). Now apply F4; all elements of K. ; ˇ/ are algebraic over K, so K. ; ˇ/ F: The elements Cˇ, ˇ, ˇ and 1= (if 0) lie in K. ; ˇ/, and thus also in F. So F really is a subfield of E. Clearly K F, since any 2 K is a zero of a polynomial X 2 KŒX and therefore algebraic over K. This completes the proof. This proof qualifies as easy, but it s only easy because we have the right notions at our disposal. Otherwise, would you be able to write down, at the drop of a hat, a nontrivial rational polynomial that vanishes at the sum of two numbers, given only rational polynomials vanishing at one and the other number respectively? F9 (Transitivity of algebraicness). Let L be an intermediate field of the extension E=K. If E=L and L=K are algebraic, so is E=K (and vice versa). Proof. Take ˇ 2 E. By assumption ˇ is algebraic over L. Let 0; 1;:::; n 1 be the coefficients of MiPo L.ˇ/; then ˇ is also algebraic over the subfield F WD K. 0; 1;:::; n 1 /. By assumption all the i are algebraic over K. Therefore we can apply F7 to conclude that F W K is finite. But F.ˇ/ W F is also finite, by F6; therefore the degree formula gives F.ˇ/WK < 1: Using F4 we see in particular that ˇ is algebraic over K.

6 20 2 Algebraic Extensions F10. Let E=K be a field extension and A a subset of E. If all elements of A are algebraic over K, the extension K.A/=K is algebraic. Proof. Clearly K.A/ is the union of all subfields of the form K.M /, wherem ranges over finite subsets of A. By F7, each K.M /=K is finite and therefore also algebraic. Thus K.A/ contains only elements algebraic over K. (Of course F10 also follows directly from F8.) F11. Let E=K be a field extension, and L 1 ; L 2 intermediate fields of E=K. The field (9) L 1 L 2 WD L 1.L 2 / D L 2.L 1 / is called the composite of L 1 and L 2 in E. (a) If L 1 =K is algebraic, so is L 1 L 2 =L 2. (b) If L 1 =K is finite, so is L 1 L 2 =L 2 ; moreover L 1 L 2 WL 2 L 1 WK. (c) If L 1 =K and L 2 =K are algebraic, so is L 1 L 2 =K. (d) If L 1 =K and L 2 =K are finite, so is L 1 L 2 =K; if, moreover, the extension degrees n 1 D L 1 WK and n 2 D L 2 WK are relatively prime, we have L 1 L 2 WK D n 1 n 2. Proof. Part (a) follows from F10, taking (9) into account. Part (c) therefore also follows, thanks to F9. Let L 1 =K and L 2 =K be finite. Assuming (b) already proved, we see from the degree formula that (10) L 1 L 2 WK D.L 1 L 2 WL 2 /.L 2 WK/.L 1 WK/.L 2 WK/; which is the first part of (d). Again from the degree formula we obtain that L 1 L 2 WK is divisible by n 1 and by n 2.Ifn 1 ; n 2 are relatively prime, L 1 L 2 WK is divisible by n 1 n 2, which together with (10) gives the second part of (d). There remains to prove (b). Consider the set R of all finite sums of products ab with a 2 L 1 ; b 2 L 2. Clearly R is a subring of E containing L 1 and L 2.Itisalso clear that any basis of L 1 =K generates R as an L 2 -vector space R, so in particular RWL 2 L 1 W K. IfL 1 WK < 1, this implies that R is a field (see F2). It follows that R D L 1 L 2, which concludes the proof.

Algebraic Cryptography Exam 2 Review

Algebraic Cryptography Exam 2 Review You should be able to do the problems assigned as homework, as well as problems from Chapter 3 2 and 3. You should also be able to complete the following exercises: